3
$\begingroup$

Let $A = F_2$ be the free group on two generators (not sure if this is important.)

Suppose you have a semidirect product $A\rtimes C$ coming from some homomorphism $\beta: C\rightarrow \text{Aut}(A)$.

Let $G$ be a finite group, and let $f_A: A\twoheadrightarrow G$ be a surjection such that $$f_A(^c\cdot) := f_A\circ\beta(c)\equiv f_A\mod\text{Inn}(G)\qquad\text{for all $c\in C$}$$

then, under these conditions, must there exist:

  1. A homomorphism $f : A\rtimes C\twoheadrightarrow G$ such that $f|_A = f_A$, or equivalently...
  2. A homomorphism $f_C : C\rightarrow G$ such that we have $$(f_A\circ\beta(c))(a) := f_A(^ca) = f_C(c)f_A(a)f_C(c)^{-1}\qquad\text{for all $c\in C, a\in A$}$$

Remark: Our assumptions tell us that for every $c\in C$, there is a $g_c\in G$ such that $f_A(^ca) = g_cf_A(a)g_c^{-1}$ for all $a\in A$, so what I'm asking is if it's always possible to pick the $g_c$'s wisely so that the map $c\mapsto g_c$ is a homomorphism.

This question came from me thinking about Teichmuller level structures and how they classify torsors, so in my context $1\rightarrow A\rightarrow A\rtimes C\twoheadrightarrow C\rightarrow 1$ is an exact sequence of fundamental groups associated to some punctured curve (either $\mathbb{P}^1$ minus 3 points or an elliptic curve punctured once).

EDIT: As Dave Witte Morris points out, this is not possible in general. Though, I wonder: Are there conditions we can put on $G$ or $f_A$ to ensure that the statement is true? If $G$ is abelian, then $f_C$ could literally be anything, but that's quite a strong condition. Can we say anything for nonabelian $G$? I'm trying to understand exactly why this fails... Is there some kind of cohomology going on?

$\endgroup$
1
  • 1
    $\begingroup$ If the center of $G$ is trivial, then $f_C$ certainly exists. In general, you have a homomorphism $\varphi$ from $C$ to $\mathrm{Inn}(G)$, and $G$ is a central extension of $\mathrm{Inn}(G)$, so you also have a cohomology class $\xi \in H^2 \bigl( \mathrm{Inn}(G); Z(G) \bigr)$. I think the obstruction to the existence of $f_C$ is the pullback $\varphi^* \xi \in H^2 \bigl( C; Z(G) \bigr)$. $\endgroup$ Sep 22, 2015 at 6:40

1 Answer 1

5
$\begingroup$

No. For a counterexample, let $G = \langle i,j \rangle$ be the quaternion group of order $8$, let $f_A$ map the generating set $\{a,b\}$ of $F_2$ to $\{i,j\}$, and let $C = \langle c \rangle$, where $c$ is the automorphism that fixes $a$ and inverts $b$.

We can choose $g_c$ to be $i$, so the hypothesis is satisfied. However, since $c$ has order 2, and the only element of order 2 in $G$ is central, there is no way to pick $g_c$ to get a homomorphism.

$\endgroup$
1
  • $\begingroup$ Ah, nice example. I'll accept this answer when I wake up tomorrow, though do you know of any conditions on $G$ or $f_A$ that would make the statement true? (Certainly if $G$ is abelian, then $f_C$ can be anything but is there anything else?) $\endgroup$
    – Will Chen
    Sep 21, 2015 at 7:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.